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Thread: Is the inverse of a function always a function ?

  1. #1
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    Is the inverse of a function always a function ?

    Hi

    Is the inverse of a function always a function ?

    I asked myself this question because of an exercise that I was solving but couldn't find an answer.
    We have [tex]f(x)=\sqrt{1+sinx}[/tex] defined over [tex]\left [\frac{-\pi}{2} ,\frac{\pi}{2} \right ][/tex]
    Let's see its curve on the graph and also the one of its inverse:

    Screenshot_20180203-201741.jpg

    The inverse's curve doesn't seem to be a function to me (maybe I'm missing some information in my mind).
    A function is a map (every x has a unique y-value), while on the inverse's curve some x-values have 2 y-values.

    By doing some calculations, I get [tex]f^{-1}(x)\,=\,\arcsin(x^{2}-1)[/tex]
    Its curve (the one I calculated) is like that (the grey one below):

    Screenshot_20180203-201851.jpg

    The part defined over negative x of this curve shocked me.
    Can someone explain to me how this can happen ?
    And why shouldn't the grey curve be like the purple one ?
    Last edited by stapel; 02-03-2018 at 03:05 PM. Reason: Copying question from subject line into post.

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Alfredo Dawlabany View Post
    Is the inverse of a function always a function?
    Obviously not! Easy case: a parabola. Its inverse is a sideways parabola, which is not a function. To be stated in functional terms, it would have to be expressed as two square-root functions.

    Quote Originally Posted by Alfredo Dawlabany View Post
    I asked myself this question because of an exercise that I was solving but couldn't find an answer.
    We have [tex]f(x)=\sqrt{1+sinx}[/tex] defined over [tex]\left[\frac{-\pi}{2} ,\frac{\pi}{2} \right][/tex]
    Let's see its curve on the graph and also the one of its inverse:

    Screenshot_20180203-201741.jpg

    The inverse's curve doesn't seem to be a function to me (maybe I'm missing some information in my mind).
    No, it is not. But you're inverting the entire original function (which you did not restrict to the stated domain), rather than a portion. This is why the arcsine function is very carefully defined on only a very limited domain.

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    Quote Originally Posted by stapel View Post
    Obviously not! Easy case: a parabola. Its inverse is a sideways parabola, which is not a function. To be stated in functional terms, it would have to be expressed as two square-root functions.


    No, it is not. But you're inverting the entire original function (which you did not restrict to the stated domain), rather than a portion. This is why the arcsine function is very carefully defined on only a very limited domain.
    Thanks, that makes sense. I don't know why I didn't think about the parabola and it's inverse which is a conic. It's more clear now.

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